I'm not sure this is how you are expected to solve this, or even if this
is the simplest method; but:

First divide through by and rearrange to get:

Now this gives:

So there exists an integer such that:

Now the LHS (left hand side) is even, so must be odd.
So let , then:

But the LHS is divisible by so the RHS must also be divisible by so must be even., so we may write it as , and then:

Now substituting this back into gives:

So we see that for any :

is a solution.

RonL

May 15th 2006, 10:33 AM

ThePerfectHacker

Use the following theorem,
Given a diophatine equation with
and is a particular solution
Then all solutions and every solution is,
for an integer
------
You have,
You can leave it the way it is but I suggest to divide by 3,
You need to find a specific solution. Which can be done with trail and error. (Otherwise you can use the Euclidean Algorithm or Coutinued fractions to get a particular solution).
We can easily see that and work.
Also which divides 13.
Thus, all solutions are,

Note it might look different from CaptainBlack's but they are both equivalent.

May 15th 2006, 12:21 PM

delpin

Thank you

For your help... most appreciated

May 28th 2006, 07:04 AM

Soroban

Hello, delpin!

This is basically Captain Black's and Hacker's solutions ... in baby-talk.